9701_w23_qp

Question 1

Topic: 23.1

Fluorine reacts with chlorine dioxide, ClO₂, as shown.
\( F_2(g) + 2ClO_2(g) \rightarrow 2FClO_2(g) \)
The rate of the reaction is first order with respect to the concentration of \( F_2 \) and first order with respect to the concentration of \( ClO_2 \). No catalyst is involved.

(a) (i) Suggest a two-step mechanism for this reaction.
(ii) Identify the rate-determining step in this mechanism. Explain your answer.

(b) When the rate of the reaction is measured in \( \text{mol dm}^{-3} \text{s}^{-1} \), the numerical value of the rate constant, \( k \), is 1.22 under certain conditions.
(i) Complete the rate equation for this reaction, stating the overall order of the reaction.

(ii) Use your rate equation in (i) to calculate the rate of the reaction when the concentrations of \( F_2 \) and \( ClO_2 \) are both \( 2.00 \times 10^{-3} \text{mol/dm}^3 \).

(c) Under different conditions, and in the presence of a large excess of \( ClO_2 \), the rate equation is as shown.
\( \text{rate} = k_1[F_2] \)
The half-life, \( t_{1/2} \), of the concentration of \( F_2 \) is 4.00s under these conditions.
(i) Calculate the numerical value of \( k_1 \), giving its units. Give your answer to three significant figures.
(ii) An experiment is performed under these conditions in which the starting concentration of \( F_2 \) is \( 0.00200 \text{mol/dm}^3 \). Draw a graph on the grid in Fig. 1.1 to show how the concentration of \( F_2 \) changes over the first 12 s of the reaction.

(iii) Use your graph in Fig. 1.1 to find the rate of the reaction when the concentration of \( F_2 \) is \( 0.00100 \text{mol/dm}^3 \). Show your working on the graph.

▶️ Answer/Explanation

Solution

(a)(i) Two possible mechanisms:

Mechanism 1:
\( F_2 + ClO_2 \rightarrow FClO_2 + F \) (Step 1)
\( ClO_2 + F \rightarrow FClO_2 \) (Step 2)

Mechanism 2:
\( F_2 + ClO_2 \rightarrow F_2ClO_2 \) (Step 1)
\( F_2ClO_2 + ClO_2 \rightarrow 2FClO_2 \) (Step 2)

Explanation: Both mechanisms are valid as they satisfy the stoichiometry of the overall reaction and the given rate orders.

(a)(ii) The first step is the rate-determining step because it involves one molecule of \( F_2 \) and one molecule of \( ClO_2 \), matching the first-order dependence on both reactants in the rate equation.

(b)(i) The rate equation is:
\( \text{rate} = k[F_2][ClO_2] \), where \( k = 1.22 \). The overall order is 2 (first-order in \( F_2 \) and first-order in \( ClO_2 \)).

(b)(ii) Substituting the given concentrations:
\( \text{rate} = 1.22 \times (2.00 \times 10^{-3}) \times (2.00 \times 10^{-3}) = 4.88 \times 10^{-6} \text{mol dm}^{-3} \text{s}^{-1} \).

(c)(i) For a first-order reaction, \( k_1 = \frac{0.693}{t_{1/2}} \).
\( k_1 = \frac{0.693}{4.00} = 0.173 \text{s}^{-1} \) (3 significant figures).

(c)(ii) The graph should show exponential decay:
– At \( t = 0 \), \([F_2] = 0.00200 \text{mol/dm}^3 \).
– At \( t = 4 \text{s} \), \([F_2] = 0.00100 \text{mol/dm}^3 \).
– At \( t = 8 \text{s} \), \([F_2] = 0.00050 \text{mol/dm}^3 \).
– At \( t = 12 \text{s} \), \([F_2] = 0.00025 \text{mol/dm}^3 \).

(c)(iii) The rate at \([F_2] = 0.00100 \text{mol/dm}^3 \) is the slope of the tangent at that point. From the graph, the rate is approximately \( 1.5-2.0 \times 10^{-4} \text{mol dm}^{-3} \text{s}^{-1} \).

Question 2

Topic: 24.2

(a) Define \(K_w\) mathematically by completing the expression.

(b) Two solutions, V and W, are described.
• V is HCl(aq).
• W is NaOH(aq).
• The concentration of HCl in V is the same as the concentration of NaOH in W.
• The pH values of V and W differ by exactly 11.00 at 298K.
(i) Calculate the concentration of HCl in V

(ii) Equal volumes of the two solutions V and W are mixed, giving solution X. Name solution X and state its pH.

(iii) A 1 cm³ sample of 1.0 mol/dm³ HNO₃ is added to 100 cm³ of solution X, forming mixture Y. A 1 cm³ sample of 1.0 mol/dm³ KOH is added to 100 cm³ of solution X, forming mixture Z. Estimate the pH of mixtures Y and Z. No calculations are required.

(c) (i) CH₃CH₂COOH, CH₃CCl₂COOH, and H₂SO₄ are all acidic. Suggest the trend in the relative acid strength of these three compounds. Explain your answer

(ii) When concentrated H₂SO₄ is added to water a series of acid-base reactions occurs. There are three conjugate acid-base pairs that can be identified during this series of reactions. Write the formulae of these three conjugate acid-base pairs.
conjugate acid 1 ………………………………….. conjugate base 1 …………………………………..
conjugate acid 2 ………………………………….. conjugate base 2 …………………………………..
conjugate acid 3 ………………………………….. conjugate base 3 …………………………………..

(d) The partition coefficient, \(K_{pc}\), of a substance, Q, between hexane and water is 7.84 at 298K. Q is more soluble in hexane than it is in water.
(i) Define partition coefficient, \(K_{pc}\).

(ii) 5.00g of Q is shaken with a mixture of 100.0 cm³ of water and 100.0 cm³ of hexane at 298K and left until there is no further change in concentrations. Calculate the mass of Q dissolved in the water.

(iii) A sample of Q is shaken with a different mixture of water and hexane and left until there is no further change in concentrations. It is found that the mass of Q dissolved in each solvent is the same. Use the \(K_{pc}\) value to suggest possible values for the volume of water used and the volume of hexane used.

(iv) Q is more soluble in hexane than it is in water. It is suggested that Q is one of KCl, CH₃(CH₂)₄OH, or HCOOH. Identify Q. Explain your answer

▶️ Answer/Explanation

Solution

(a) \(K_w = [H^+][OH^-]\) or \(K_w = [H_3O^+][OH^-]\)

Explanation: The ion product of water (\(K_w\)) is the product of the concentrations of \(H^+\) and \(OH^-\) ions in water at equilibrium, which is \(1.0 \times 10^{-14}\) at 298K.

(b)(i) Concentration of HCl = 0.0316 mol/dm³

Explanation: Given the pH difference is 11, the pH of HCl (V) is 1.5 and NaOH (W) is 12.5. Using \([H^+] = 10^{-pH}\), \([H^+] = 10^{-1.5} = 0.0316 \, \text{mol/dm}^3\).

(b)(ii) Solution X is sodium chloride (NaCl) with pH = 7.

Explanation: Equal volumes of strong acid (HCl) and strong base (NaOH) of the same concentration neutralize each other, forming neutral NaCl solution.

(b)(iii) Mixture Y: pH 1-3; Mixture Z: pH 11-13.

Explanation: Adding HNO₃ (strong acid) lowers pH, while adding KOH (strong base) raises pH. The exact values depend on dilution effects.

(c)(i) Acid strength: \(H_2SO_4 > CH_3CCl_2COOH > CH_3CH_2COOH\).

Explanation: \(H_2SO_4\) is a strong acid (fully dissociated). The chlorine atoms in \(CH_3CCl_2COOH\) withdraw electron density, stabilizing the conjugate base more than the alkyl group in \(CH_3CH_2COOH\).

(c)(ii) Conjugate acid-base pairs:
• \(H_3O^+\) and \(H_2O\)
• \(H_2SO_4\) and \(HSO_4^-\)
• \(HSO_4^-\) and \(SO_4^{2-}\)

Explanation: These pairs arise from the stepwise dissociation of \(H_2SO_4\) in water.

(d)(i) Partition coefficient (\(K_{pc}\)) is the ratio of solute concentrations in two immiscible solvents at equilibrium.

(d)(ii) Mass of Q in water = 0.566 g.

Explanation: Let \(x\) be mass in water, then \(\frac{5-x}{x} = 7.84\). Solving gives \(x = 0.566 \, \text{g}\).

(d)(iii) Example volumes: 78.4 cm³ water and 10 cm³ hexane.

Explanation: For equal masses, \(\frac{V_{\text{water}}}{V_{\text{hexane}}} = 7.84\) since \(K_{pc} = 7.84\).

(d)(iv) Q is \(CH_3(CH_2)_4OH\).

Explanation: It has a non-polar hydrocarbon chain, making it more soluble in hexane than water, unlike the ionic KCl or polar HCOOH.

Question 3

Topic: 25.2

Hydrogen peroxide is a liquid at 298K. It is moderately stable under room conditions but will decompose quickly if a catalyst is added.
reaction 1 \(2H_2O_2(l) \to 2H_2O (l) + O_2 (g)\)
(a) (i) Define entropy.
(ii) Predict the sign of the standard entropy change of reaction 1. Explain your answer.
(b) Some bond energy data are shown in Table 3.1.

Use the data in Table 3.1 to show that the enthalpy change of the following reaction is \(–196 kJ mol^{–1}\).
2H₂O₂(g) \(\to\) 2H₂O(g) + O₂(g)

The enthalpy change and Gibbs free energy change for the following reaction are shown.
2H₂O₂(l) → 2H₂O(l) + O₂(g) \(ΔH^o\) = –196 kJ mol⁻¹
\(ΔG^o = –238kJmol^{–1}\)
Use the data given to calculate the standard entropy of oxygen, \(S^o\) , O₂(g).

(d) The decomposition of H₂O₂(aq) is catalysed by aqueous iron(III) chloride and by silver metal. Identify which of these two catalysts is acting as a homogeneous catalyst. Explain your answer.

(e) The E o values for two electrode reactions are given.
\(H_2O_2 + 2H^+ + 2e^− \rightleftharpoons 2H_2O\) \(E^o = +1.77V\)
\(Cr^{3+} + e^− \rightleftharpoons Cr^{2+}\) \(E ^o = −0.41V\)
(i) An electrochemical cell is constructed with the following half-cells (electrodes):
• an acidified solution of H₂O₂, a platinum wire
• Cr²⁺ mixed with Cr³⁺, a platinum wire.
Identify the positive half-cell and calculate the standard cell potential, \(E^o_{cell}\).
(ii) Calculate the value of ΔG o for the cell reaction that occurs, per mole of H₂O₂.

(f) The \(E^ o\) values for two electrode reactions are given.
\(H_2O_2 + 2H^+ + 2e^− \rightleftharpoons 2H_2O\) \(E^o = +1.77V\)
\(Co^{3+} + e^– \rightleftharpoons Co^{2+}\) \(E ^o = +1.82V\)
An electrochemical cell is constructed with the following half-cells.
half-cell 1 an acidified solution of H₂O₂ under standard conditions, a platinum wire
half-cell 2 a solution containing 0.020 moldm⁻³ Co³⁺ and 2.0 moldm⁻³ Co²⁺, a platinum wire
(i) Use the Nernst equation to calculate the value of E, the electrode potential of half-cell 2 under these conditions.

(ii) Write an equation for the cell reaction that occurs in this cell under these condition.

(g) (i) Define enthalpy change of hydration, \(\Delta H_{hyd}\).
(ii) Aluminium fluoride, AlF₃, is an ionic solid.
Complete and label the energy cycle to show the relationship between:
● the enthalpy change of solution of AlF₃, \(\Delta H ^o_{sol}\)
● the lattice energy of AlF₃, \(\Delta H ^o_{latt}\)
● the enthalpy changes of hydration of Al³⁺ and F⁻, \(\Delta H^o_{hyd}\). Include state symbols for all substances and ions.

(iii) Relevant data for this question are given.

Use these data and your energy cycle in (g)(ii) to calculate the \(\Delta H ^o_{latt}\) of \(AlF_3\).

▶️ Answer/Explanation

Solution

(a)(i) Entropy is a measure of the disorder or randomness in a system, representing the number of possible arrangements of particles and energy.

(a)(ii) The sign of the standard entropy change (\(ΔS^o\)) is positive because the reaction produces a gas (\(O_2\)) from a liquid (\(H_2O_2\)), increasing the disorder in the system.

(b) Using bond energy data:
\(ΔH = (2 × 150) – (1 × 496) = -196 \, \text{kJ mol}^{-1}\). This confirms the enthalpy change for the reaction.

(c) Using \(ΔG = ΔH – TΔS\):
\(-238 = -196 – 298ΔS\) → \(ΔS = 0.141 \, \text{kJ K}^{-1} \text{mol}^{-1}\).
Then, \(S^o(O_2(g)) = 205 \, \text{J K}^{-1} \text{mol}^{-1}\).

(d) Iron(III) chloride (\(FeCl_3\)) is the homogeneous catalyst because it is in the same aqueous phase as the reactants.

(e)(i) The positive half-cell is the hydrogen peroxide electrode.
\(E^o_{cell} = 1.77 – (-0.41) = +2.18 \, \text{V}\).

(e)(ii) \(ΔG^o = -nFE^o_{cell} = -2 × 96500 × 2.18 = -420.7 \, \text{kJ mol}^{-1}\).

(f)(i) Using the Nernst equation:
\(E = 1.82 + 0.059 \log(0.02 / 2) = +1.702 \, \text{V}\).

(f)(ii) The cell reaction is:
\(H_2O_2 + 2H^+ + 2Co^{2+} → 2H_2O + 2Co^{3+}\).

(g)(i) Enthalpy change of hydration (\(ΔH_{hyd}\)) is the energy change when one mole of gaseous ions dissolves in water to form an aqueous solution.

(g)(iii) Using the energy cycle:
\(ΔH_{latt} = -4690 + (3 × -506) – (-209) = -5999 \, \text{kJ mol}^{-1}\).

Question 4

Topic: 28.4

(a) Cobalt(II) nitrate, Co(NO₃)₂, is a reddish-brown crystalline solid. It dissolves in water to form a solution containing [Co(H₂O)₆]²⁺ complex ions.
(i) Complete Table 4.1 giving the formula of the cobalt-containing species that is formed in each of the three reactions described.

(ii) Describe the colour change seen in reaction 3.
original colour of [Co(H₂O)₆]²⁺(aq) …………………..
final colour after addition of an excess of conc. HCl(aq) …………………………
(b) Calcium nitrate, Ca(NO₃)₂, is a white crystalline solid. When heated, it starts to decompose at approximately 500°C.
(i) Write an equation for the decomposition of Ca(NO₃)₂.
(ii) Suggest temperatures at which Mg(NO₃)₂ and Ba(NO₃)₂ start to decompose. Explain your answer.

▶️ Answer/Explanation

Solution

(a)(i)

Answer:
• Co(OH)₂ OR Co(OH)₂(H₂O)₄
• \([Co(NH_3)_6]^{2+}\) OR \([Co(NH_3)_6]^{3+}\)
• [CoCl₄]²⁻

Explanation: In reaction 1, NaOH precipitates Co(OH)₂. In reaction 2, NH₃ forms \([Co(NH_3)_6]^{2+}\). In reaction 3, HCl forms the blue \([CoCl_4]^{2-}\) complex.

(a)(ii)

Answer: pink to blue

Explanation: The \([Co(H_2O)_6]^{2+}\) ion is pink, while the \([CoCl_4]^{2-}\) ion formed in excess HCl is blue.

(b)(i)

Answer: \(2Ca(NO_3)_2 \rightarrow 2CaO + 4NO_2 + O_2\) OR \(Ca(NO_3)_2 \rightarrow CaO + 2NO_2 + \frac{1}{2}O_2\)

Explanation: Thermal decomposition of Ca(NO₃)₂ produces CaO, NO₂, and O₂ as the products.

(b)(ii)

Answer:
• Mg(NO₃)₂ decomposes below 480°C
• Ba(NO₃)₂ decomposes above 520°C

Explanation: Larger cations (Ba²⁺) have lower charge density and polarize the nitrate ion less, requiring higher decomposition temperatures compared to smaller cations (Mg²⁺).

Question 5

Topic: 28.3

Transition elements behave as catalysts and can form complex ions.

(a) Explain why transition elements behave as catalysts.

(b) Silver forms the linear complex ion \([Ag(CN)_2]^–\). Copper forms the tetrahedral complex ion \([Cu(CN)_4]^{3−}\). Titanium forms the complex [TiCl₄(diars)₂], where diars is a neutral bidentate ligand.
(i) State the oxidation state and the coordination number of titanium in [TiCl₄(diars)₂].
(ii) Draw three-dimensional diagrams to show the shapes of [Ag(CN)₂]⁻ and [Cu(CN)₄]³⁻ in the boxes. Label one bond angle on each diagram.

(c) The numerical value of the stability constant, Kstab, of the copper(I) complex [Cu(CN)₄]³⁻ is \(2.0 × 10^{27}\).
(i) Write an expression for the \(K_{stab}\) of [Cu(CN)₄]³⁻.

(ii) In a solution the concentrations of CN⁻ and [Cu(CN)₄]³⁻ are both 0.0010 mol/dm³. Use your expression from (c)(i) and the value of \(K_{stab}\) to calculate the concentration of \(Cu^+\)(aq) in this solution.

(d) A piece of a copper-containing alloy has a mass of 0.567g. It is dissolved in an acid giving 100.0 cm³ of a blue solution in which all the copper is present as Cu²⁺ ions. An excess of KI(aq) is added to a 25.0 cm³ sample of this solution. All of the copper is precipitated as white CuI(s). Cu²⁺ ions are the only component in the solution that react with KI(aq). This is reaction 1.
reaction 1 \(2Cu^{2+} + 4I^−\to 2CuI + I_2\)
The liberated I₂ is then titrated with 0.0200 moldm⁻³ S₂O₃²⁻. This is reaction 2.
reaction \(2 I_2 + 2S_2O_3^{2–} \to 2I^− + S_4O_6^{2–}\)
The titration requires 20.10 cm³ of 0.0200 mol/dm³ S₂O₃²⁻ to reach the end-point.
(i) Calculate the number of moles of I₂ that are reduced in this titration.

(ii) Calculate the number of moles of copper in the original piece of alloy.

(iii) Calculate the percentage of copper in the alloy.

(iv) Suggest why a solution of Cu²⁺ is coloured but solid CuI is white.

▶️ Answer/Explanation

Solution

(a) Transition elements behave as catalysts because they have variable oxidation states and vacant d-orbitals. These properties allow them to form intermediate complexes and facilitate reactions by lowering activation energy.

Explanation: The ability to change oxidation states enables transition metals to participate in redox reactions, while their vacant d-orbitals can accept electron pairs from reactants, stabilizing transition states.

(b)(i) Oxidation state = +4, Coordination number = 8.

Explanation: In [TiCl₄(diars)₂], Ti forms bonds with 4 Cl⁻ (each contributing -1) and 2 neutral diars ligands. The oxidation state is +4, and the coordination number is 8 (4 from Cl⁻ and 4 from the two bidentate diars ligands).

(b)(ii)

Explanation: [Ag(CN)₂]⁻ is linear (bond angle = 180°), while [Cu(CN)₄]³⁻ is tetrahedral (bond angle ≈ 109.5°). The diagrams show these shapes with labeled angles.

(c)(i) \(K_{stab} = \frac{[[Cu(CN)_4]^{3–}]}{[Cu^+][CN^–]^4}\)

Explanation: The stability constant expression is derived from the equilibrium concentrations of the complex ion and its constituent ions.

(c)(ii) \([Cu^+] = 5.0 × 10^{-19} \, \text{mol dm}^{-3}\)

Explanation: Substituting the given values into the \(K_{stab}\) expression: \([Cu^+] = \frac{0.0010}{2.0 × 10^{27} × (0.0010)^4} = 5.0 × 10^{-19}\).

(d)(i) Moles of I₂ = \(2.01 × 10^{-4}\)

Explanation: From the titration, moles of S₂O₃²⁻ = \(0.0200 × 0.02010 = 4.02 × 10^{-4}\). Since 2 moles of S₂O₃²⁻ react with 1 mole of I₂, moles of I₂ = \(2.01 × 10^{-4}\).

(d)(ii) Moles of Cu = \(1.61 × 10^{-3}\)

Explanation: From reaction 1, 2 moles of Cu²⁺ produce 1 mole of I₂. Thus, moles of Cu²⁺ in 25 cm³ = \(4.02 × 10^{-4}\). Scaling up to 100 cm³ gives \(1.61 × 10^{-3}\) moles.

(d)(iii) Percentage of Cu = 18.0%

Explanation: Mass of Cu = \(1.61 × 10^{-3} × 63.5 = 0.102 \, \text{g}\). Percentage = \(\frac{0.102}{0.567} × 100 = 18.0\%\).

(d)(iv) Cu²⁺ has a partially filled d-subshell, allowing d-d transitions that absorb visible light, giving color. CuI has a full d-subshell (d¹⁰), so no electronic transitions occur in the visible range, making it white.

Question 6

Topic: 28.5

(a) Five ligands are listed in Table 6.1.

(i) Complete Table 6.1 using the words monodentate, bidentate and polydentate only. Each of these three words may be used once, more than once, or not at all.
(ii) The molecule H₂NCH₂CH₂NHCH₂CH₂NH₂ is a tridentate ligand. Suggest the meaning of tridentate ligand.
(iii) Suggest how H₂NCH₂CH₂NHCH₂CH₂NH₂ acts as a tridentate ligand.
(b) Nickel forms the octahedral complex \([Ni(en)_2(H_2O)_2]^{2+}\). This complex can exist in three isomeric forms, listed in Table 6.2. One of these forms is a trans isomer, the other forms are two different cis isomers.

(i) Complete Table 6.2 using the terms polar or non-polar. Each term may be used once, more than once, or not at all.
(ii) Describe the difference between cis isomer 1 and cis isomer 2.

▶️ Answer/Explanation

Solution

(a)(i)

NH₃ = monodentate
\(EDTA^{4–}\) = polydentate / hexadentate
CN⁻ = monodentate
C₂O₄²⁻ = bidentate

Explanation: NH₃ and CN⁻ are monodentate as they donate one lone pair. C₂O₄²⁻ is bidentate due to two oxygen donor atoms, while \(EDTA^{4–}\) is polydentate as it has multiple donor sites.

(a)(ii) (ligand that) donates 3 lone pairs to central metal atom / ion OR forms 3 dative bonds to central metal atom / ion.

Explanation: A tridentate ligand binds to the metal center using three donor atoms, forming three coordinate bonds.

(a)(iii) Uses the lone pairs on each of the three nitrogen atoms to form dative bonds with the metal ion.

Explanation: The three amine groups (–NH₂ and –NH–) in the ligand each provide a lone pair to coordinate with the metal ion.

(b)(i)

trans = non-polar
cis isomer 1 = polar
cis isomer 2 = polar

Explanation: The trans isomer is non-polar due to symmetric charge distribution, while both cis isomers are polar because of asymmetric ligand arrangement.

(b)(ii) Optical isomers / non-superimposable mirror images.

Explanation: The two cis isomers are enantiomers, meaning they are mirror images that cannot be superimposed, rotating plane-polarized light in opposite directions.

Question 7

Topic: 35.2

Sunset Yellow is an additive used for colouring foods. A synthetic route for making Sunset Yellow is shown. Molecules E and G each contain one \(–SO_3^–\) group. These groups are unchanged in the formation of Sunset Yellow.

(a) State the molecular formula of the Sunset Yellow anion.
(b) Deduce the structures of E, F and G and draw them in the boxes in Fig. 7.1.
(c) Suggest suitable reagents and conditions for step 1 and 2.
(d) Predict the number of peaks in the carbon-13 NMR spectrum of the Sunset Yellow anion.

▶️ Answer/Explanation

Solution

(a) \(C_{16}H_{10}N_2O_7S_2^{2–}\) OR \(C_{16}H_{10}N_2O_7S_2\)

Explanation: The molecular formula is derived by counting all carbon, hydrogen, nitrogen, oxygen, and sulfur atoms in the Sunset Yellow anion structure, including the two \(–SO_3^–\) groups.

(b)

Explanation: Structures E, F, and G are deduced based on the given synthetic route. E and G retain their \(–SO_3^–\) groups, while F is the intermediate diazonium salt formed during the reaction.

Explanation: Step 1 involves diazotization, requiring nitrous acid (HNO₂) or NaNO₂ + HCl at low temperature (≤10°C). Step 2 is an azo coupling reaction under alkaline conditions (NaOH) to form the final product.

(d) 14 / fourteen

Explanation: The Sunset Yellow anion has 14 unique carbon environments, as each carbon in the aromatic rings and functional groups gives a distinct peak in the carbon-13 NMR spectrum.

Question 8

Topic: 34.4

Capsaicin is found in chilli peppers.

You should assume the CH₃O group is unreactive in the reactions involved in this question.
(a) Name all the functional groups in capsaicin in addition to the CH₃O group.
(b) Complete the equation for the reaction of capsaicin with an excess of Br₂(aq) in the dark. Draw the structure of the organic product in the labelled box.

(c) Capsaicin is heated with an excess of hydrogen gas in the presence of platinum metal. The six-membered ring reacts in the same way as benzene under these conditions. Draw the structure of the organic product formed.

(d) When capsaicin is treated with reagent J under suitable conditions one of the products is methylpropanoic acid, CH₃CH(CH₃)COOH.
(i) Identify reagent J and any necessary conditions.

(ii) There are three different peaks in the proton (1H) NMR spectrum of CH₃CH(CH₃)COOH in CDCl₃.

(e) (i) Capsaicin is heated with an excess of hot aqueous NaOH

(ii) Name the two types of reaction occurring in (e)(i).
(f) Draw the structure of the organic product L formed when capsaicin is treated with \(LiAlH_4\) in dry ether.

▶️ Answer/Explanation

Solution

(a) phenol, amide AND alkene / C=C all three needed [1]

Explanation: The functional groups in capsaicin (excluding the unreactive CH₃O group) are the phenol (hydroxyl attached to a benzene ring), amide (peptide-like linkage), and alkene (C=C double bond).

(b)

Explanation: Bromine (Br₂) reacts with the alkene (C=C) in capsaicin to form a dibromo compound. The product retains all other functional groups since they are unreactive under these conditions.

(c)

Explanation: Hydrogenation with Pt reduces the benzene ring (to cyclohexane) and the C=C bond (to C-C), while the amide and phenol groups remain unchanged.

(d)(i) hot AND concentrated acidified AND MnO₄⁻ / KMnO₄ all [1]

Explanation: Oxidative cleavage of the alkene side chain requires hot, concentrated KMnO₄ under acidic conditions to yield methylpropanoic acid.

(d)(ii)

Explanation: The three NMR peaks correspond to the CH₃ (methyl), CH (methine), and COOH (carboxylic acid) protons in methylpropanoic acid.

(e)(i)

(e)(ii) hydrolysis AND neutralisation / acid-base [1]

Explanation: The amide undergoes hydrolysis (breaking into amine and carboxylate), followed by neutralisation of the carboxylate to form a carboxylic acid.

(f)

Explanation: \(LiAlH_4\) reduces the amide to an amine and the phenol to an alcohol, while leaving the alkene and benzene ring intact.

Question 9

Topic: 34.4

(a) Benzoyl chloride, C₆H₅COCl, can be made from ethyl benzene in a two-step process. A reaction scheme is shown.

(i) Draw the intermediate organic compound M in the box.
(ii) Suggest suitable reagents and conditions for step 1 and step 2.
(iii) Identify the type of reaction in step 1 and step 2.

(b) C₆H₅COCl reacts with phenol, C₆H₅OH, to give the ester phenyl benzoate, C₆H₅COOC₆H₅. An incomplete description of the mechanism of this reaction is shown in Fig. 9.1.

(i) Complete the mechanism in Fig. 9.1 and include:
● all relevant dipoles (δ+ and δ–) and full electric charges (+ and –) on the species in box one and in box two
● all relevant lone pairs on the species in box one and in box two
● all relevant curly arrows to show the movement of electron pairs in box one and in box two
● the formula of the second product in box three.
(ii) Name this mechanism.
(c) Benzoyl chloride, chlorobenzene and chloroethane differ in their rates of hydrolysis when each compound is added separately to water at 25°C. Suggest the relative ease of hydrolysis of these three compounds. Explain your answer.

▶️ Answer/Explanation

Solution

(a)

Explanation: The intermediate M is benzoic acid (C₆H₅COOH), formed by oxidation of ethyl benzene. Step 1 uses hot alkaline KMnO₄ for oxidation, and Step 2 uses SOCl₂ (or PCl₅/PCl₃) for nucleophilic substitution to convert the carboxylic acid to benzoyl chloride.

(b)

Explanation: The mechanism is nucleophilic addition-elimination. The phenol’s lone pair attacks the carbonyl carbon (δ+), forming a tetrahedral intermediate. The C-Cl bond breaks, releasing Cl⁻, and the C=O reforms, producing phenyl benzoate and HCl.

(c) Order of hydrolysis ease: Benzoyl chloride > Chloroethane > Chlorobenzene.

Explanation: Benzoyl chloride hydrolyzes fastest due to the electron-withdrawing C=O group making the C-Cl bond weaker. Chloroethane hydrolyzes slower due to the alkyl group’s electron-donating effect. Chlorobenzene is the slowest because the C-Cl bond has partial double-bond character due to resonance.

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